Explanation:
The given data is as follows.
Area (A) = 0.7 [tex]m^{2}[/tex].
Electric field between the plates, (E) = 55 N/C
Since, electric field is related to surface charge density as follows.
[tex]\sigma = \frac{Q}{A}[/tex]
Also, E = [tex]\frac{\sigma}{\epsilon_{o}}[/tex]
or, [tex]\sigma = \epsilon_{o} E[/tex]
Therefore, charge will be calculated as follows.
Q = [tex]\sigma \times A = \epsilon_{o} E \times A[/tex]
= [tex]8.854 \times 10^{-12} \times 55 \times 0.7[/tex]
= [tex]3.41 \times 10^{-10} C[/tex]
= 341 pC
Thus, we can conclude that magnitude of the charge on each plate is 341 pC.
